Question

Evaluate the triple integral. X dv, where e is bounded by the paraboloid x = 4y2 4z2 and the plane x = 4. E

Answer 1

The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as 16.762

Parabloid, x = 4y² + 4z²

plane x = 4

x = 4y² + 4z²

x = 4

4 = 4y² + 4z²

4 = 4 (y² + z² )

1 = y² + z²

from polar coordinates

y = r cos θ

z = r sin θ

r² = y² + z²

The limits of the integral

0 ≤ θ ≤ 2π

4r² ≤ x ≤ 4

0 ≤ r ≤ 1

$\int\limits\int\limits\int\limits {x} \, dV = \int\limits^1_0\int\limits^a_b\int\limits^c_d {x} \, dx ( rdrdz)$

where

a = 4

b = 4r²

c = 2r

d = 0

The first integral using limits c and d gives:

$2pi\int\limits^1_0\int\limits^a_b {xr} \, dx$

The second integral using limits a and b

$pi\int\limits^1_0 {16 } } \, rdr - pi\int\limits^1_0 {16r^{5} \, dx$

$16pi\int\limits^1_0 { } } \, rdr - 16pi\int\limits^1_0 {r^{5} \, dx$

$16pi\int\limits^1_0 { } } \, [r-r^{5}]dr$

The third integral using limits 1 and 0 gives: 16.762

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Answer 2

The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as 16.762

What is integration?

Integration is defined as adding small parts to form a new significant part.

Parabloid, x = 4y² + 4z²

plane x = 4

x = 4y² + 4z²

x = 4

4 = 4y² + 4z²

4 = 4 (y² + z² )

1 = y² + z²

from polar coordinates

y = r cos θ

z = r sin θ

r² = y² + z²

The limits of the integral

0 ≤ θ ≤ 2π

4r² ≤ x ≤ 4

0 ≤ r ≤ 1

$\int\int\intxdV = \int_0_1\int_b_a\int_d_cxdx(rdrdz)$

where

a = 4

b = 4r²

c = 2r

d = 0

The first integral using limits c and d gives:

$2\pi\int_0^1\int_b^axydx$

The second integral using limits a and b

$\pi \int_0^116rdr-\pi\int_0^116r^5dx$

$16\pi \int_0^1rdr-16\pi\int_0^1 r^5dx$

$16\pi\int_0^1[r-r^5]dr$

The third integral using limits 1 and 0 gives: 16.762

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